(v) (−52,π3)(-\dfrac{5}{2},\dfrac{\pi}{3})(−25,3π) Here r=−52<0r=-\dfrac{5}{2}<0r=−25<0 and θ=π3\theta=\dfrac{\pi}{3}θ=3π. Since r<0r<0r<0, use angle: π3+π=4π3\frac{\pi}{3}+\pi=\frac{4\pi}{3}3π+π=34π x=−52cosπ3=−54y=−52sinπ3=−534\begin{aligned} x&=-\frac{5}{2}\cos\frac{\pi}{3}=-\frac{5}{4}\\ y&=-\frac{5}{2}\sin\frac{\pi}{3}=-\frac{5\sqrt{3}}{4} \end{aligned}xy=−25cos3π=−45=−25sin3π=−453